If in a circuit a resistor with given value of R should be of W watt, is it OK to replace it with following cases:

2 \$R/2\$ of \$W/2\$ in series.

2 \$R\cdot2\$ of \$W/2\$ in parallel.

In other words, can we increase heat dissipation by putting resistors in series or parallel? what are the pros and cons if any?


Both series and parallel connection can work, with suitable changes to the resistor values as Anindo described..

Putting resistors in series has the feature that if one burns out (fails open-circuit) you lose all the load. In parallel, if one burns out, you lose half the load (or less, if there are more that 2 resistors in parallel.)

There are circumstances where you want an immediate and total shutdown after a failure - then series connection is best.

There are probably more circumstances where a graceful degradation is better - slow braking a motor is better than none! Dim LEDs are probably better than darkness... Then, parallel connection is best.

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  • 1
    \$\begingroup\$ Is it your experience that resistors trying to dissipate too much heat generally fail open? I've never really thought about it but for some reason my assumption was they would fail closed from melting or something... \$\endgroup\$ – NickHalden Apr 29 '13 at 23:36
  • \$\begingroup\$ Pretty much all failed resistors I have come across were open-circuit. However, if you design safety-critical stuff, you must assume short-circuits, too. For instance, a solder blob under a 0402 SMD resistor is not too uncommon. \$\endgroup\$ – zebonaut Apr 30 '13 at 7:45
  • \$\begingroup\$ I can't recall seeing a resistor fail short in nearly 40 years, but Zebonaut's point MUST be borne in mind for safety-critical work. (Even that isn't a resistor failing and could apply to any overheating fine pitch component.) Diodes, now, and transistors, and some capacitors, can fail short. \$\endgroup\$ – Brian Drummond Apr 30 '13 at 9:34

In order to "share the load" between resistors, each resistor must be able to dissipate half the desired wattage.

In the case of the question, if the desired resistance is R and wattage W, then putting two (ideal) resistors of value 2 x R each in parallel will result in a resistance of R, but with each of the resistors passing only half of the current.

For a parallel combination of resistors, the resultant resistance is given by:
R = 1 / ( 1/R1 + 1/R2)

Thus, to achieve a value of R using a parallel combination of two resistors, we must use R1 = R2 = 2 x R

Now to calculate the power:
Power P = V^2 / R

The the voltage across the resistor combination is constant, V
Power through resistor R1 from above: = P1 = V^2 / (2 x R) = P / 2
Similarly P2 = V^2 / (2 x R) = P / 2

Thus, each resistor will dissipate half the specified power, or in other words will need to be half the stated power rating W.

In reality, resistors differ in actual resistance due to manufacturing tolerances. Hence, one resistor might carry slightly more current than another, both being nominally of equal value. To provide headroom for such variations, it is best to allow for slightly higher power ratings for the resistors, than is absolutely required.

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